During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 g mL^(-1) to 1.139 g mL^(-). Sulphuric acid of density 1.294 g mL^(-1) is 39% by weight and that of density 1.139 g mL^(-1) is 20% by weight. The battery hold 3.5 litre of acied and discharge. Calculate the no. of ampere hour for which the battery must have been used. The charging and discharging reactions are: Pb+SO_(4)^(2-) rarr PbSO_(4)+2e (charging) PbO_(2)+4H^(+)+SO_(4)^(2-)+2e rarr PbSO_(4)+2H_(2)O (discharging)

During the discharge of a lead storage battery, the density of sulphur

1.	During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 g mL^(-1) to 1.139 g mL^(-). Sulphuric acid of density 1.294 g mL^(-1) is 39% by weight and that of density 1.139 g mL^(-1) is 20% by weight. The battery hold 3.5 litre of acied and discharge. Calculate the no. of ampere hour for which the battery must have been used. The charging and discharging reactions are: Pb+SO_(4)^(2-) rarr PbSO_(4)+2e (charging) PbO_(2)+4H^(+)+SO_(4)^(2-)+2e rarr PbSO_(4)+2H_(2)O (discharging)
Answer» Solution :The overall battery reaction is `Pb+PbO_(2) + 2H_(2)SO_(4) = 2PbSO_(4) + 2H_(2)O` `because` two moles of electrons are involved for the reaction of two moles `H_(2)SO_(4)`. `THEREFORE` eq. wt. of `H_(2)SO_(4)` = mol. Wt. of `H_(2)SO_(4) = 98`. Following in the same WAY as in no. of eq. of `H_(2)SO_(4)` present in 3.5 litres of solution of a charged battery `= (39)/(98) xx (1.294)/(100) xx 3500` `= 18.0235` No. of equivalents of `H_(2)SO_(4)` present in 3.5 litres of solution after getting discharged `= (20)/(98) xx (1.139)/(100) xx 3500` = 8.1357 Number of eq. of `H_(2)SO_(4)` lost = 18.0235 - 8.1357 = 9.8878 `therefore` moles of electric charge produced by the battery = 9.8878 F `= 9.8878 xx 96500` coulombs `= 9.8878 xx 96500` amp - seconds `= (9.8878 xx 96500)/(60 xx 60)` amp - HOURS =265 amp - hours.